A geometry problem by Krishna Shankar

Geometry Level 4

2 sin x + 3 sin x + 6 sin x = 1 \large 2^{\sin{x}} + 3^{\sin{ x}} + 6^{\sin {x}} = 1

Find the smallest positive integer x x satisfying the equation above.

Clarification: Angles are measured in degrees.


Inspiration .


The answer is 270.

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1 solution

Rishabh Jain
Dec 15, 2016

Note 1 sin x 1 x R -1\le\sin x\le 1\forall x\in\mathfrak R . Also since a x a^x is an increasing function for real a 'a' , we can write a x [ a 1 , a 1 ] a^x\in [a^{-1},a^{1}] . Hence 2 x + 3 x + 6 x [ 2 1 + 3 1 + 6 1 , 2 + 3 + 6 ] = [ 1 , 11 ] 2^x+3^x+6^x\in\color{#D61F06}{[}2^{-1}+3^{-1}+6^{-1},2+3+6]=\color{#D61F06}{[}1,11] . Hence according to the question we have the case when minimum value of expression occurs ie when sin x = 1 \sin x=-1 and least integer satisfying this criteria is x = 27 0 x=270^{\circ} .

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